The given expression is the limit as \( x \to 0 \) of \( \frac{\sin(x)}{\sqrt{\log(1+x)}} \):

\[
\lim_{x \to 0} \frac{\sin(x)}{\sqrt{\log(1+x)}}
\]

To solve this, let's first analyze the behavior of both the numerator and the denominator as \( x \to 0 \).

1. **Numerator**: As \( x \to 0 \),
   \[
   \sin(x) \approx x \quad \text{(since \( \sin(x) \) behaves like \( x \) near zero)}.
   \]
   So, \( \sin(x) \to 0 \) as \( x \to 0 \).

2. **Denominator**: For small values of \( x \), we can approximate \( \log(1+x) \) using the first term in its Taylor expansion:
   \[
   \log(1+x) \approx x \quad \text{(since \( \log(1+x) \approx x \) for small \( x \))}.
   \]
   Hence,
   \[
   \sqrt{\log(1+x)} \approx \sqrt{x}.
   \]

Now the expression becomes:

\[
\lim_{x \to 0} \frac{x}{\sqrt{x}} = \lim_{x \to 0} \sqrt{x}.
\]

As \( x \to 0 \), \( \sqrt{x} \to 0 \).

Thus, the limit of the given expression as \( x \to 0 \) is:

\[
\boxed{0}.
\]

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. What is the Taylor expansion of \( \sin(x) \)?
2. How can L'Hopital's rule be applied to limits like this one?
3. What are the general properties of logarithmic functions near zero?
4. Can you approximate higher-order terms in the expansion of \( \log(1+x) \)?
5. What is the limit of \( \frac{\sin(x)}{\log(1+x)} \) as \( x \to 0 \)?

**Tip**: In many limits involving trigonometric and logarithmic functions, approximations using Taylor series are useful to simplify the expressions.

Explore the limit of the expression sin(x)/sqrt(log(1+x)) as x approaches 0 using Taylor series approximation. This detailed solution breaks down the behavior of trigonometric and logarithmic functions, providing insights into key calculus concepts. Suitable for undergraduate math students.

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